Question

# Find the moment of inertia of a cylinder of mass $$M$$, radius $$R$$ and length $$L$$ about an axis passing through its centre and perpendicular to its length.

Solution

## Moment of inertia of disc $$=\dfrac{MR^2}{2}$$Moment of inertia of disc parallel to its plane $$=\dfrac{MR^2}{4}$$   ( by perpendicular anus theorem ).For a small part of length $$dx$$ amount of inertia will $$=\dfrac{dmR^2}{4}+dm\;x^2$$   ( by parallel axis theorem )$$\Rightarrow I=\displaystyle\int _{ -L/2 }^{ L/2 }{ dm\left( \dfrac { { R }^{ 2 } }{ 4 } +{ x }^{ 2 } \right) }$$        $$=\displaystyle \int _{ -L/2 }^{ +L/2 }{ \dfrac { M }{ L } dx\left( \dfrac { { R }^{ 2 } }{ 4 } +{ x }^{ 2 } \right) }$$        $$={ \left[ \dfrac { M }{ L } \left( \dfrac { { R }^{ 2 } }{ 4 } x+\dfrac { { x }^{ 3 } }{ 3 } \right) \right] }_{ -L/2 }^{ L/2 }$$        $$=\dfrac { M }{ L } \left( \dfrac { { R }^{ 2 } }{ 4 } L+\dfrac { { L }^{ 3 } }{ 12 } \right)$$        $$=\dfrac { M }{ 4 } \left( { R }^{ 2 }+\dfrac { { L }^{ 2 } }{ 3 } \right)$$Hence, the answer is $$\dfrac { M }{ 4 } \left( { R }^{ 2 }+\dfrac { { L }^{ 2 } }{ 3 } \right) .$$Physics

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